Kinetics of the Formation of HBr

Kinetics of the Formation of HBr

Kinetics of the Formation of HBr


Kinetics of the Photochemical Reaction Between Hydrogen and Bromine

The photochemical combination of moist H2 and Br2 vapor in visible light(< 510nm) is a chain reaction and is occurs at 433-491 k.
H2 + Br2 ---h𝜈→ 2HBr
Possible mechanism of this photochemical reaction-
Chain Initiation-
1. Br2 + h𝜈 ---k1→ 2Br
Chain Propagation-
ii. Br + H2 ---k2→ HBr + H
iii. H + Br2 ---k3→ HBr + Br
Chain Inhibition-
iv. H + HBr ---k4→ H2 + Br
Chain Termination-
v. Br + Br ---k5→ Br2
where, k1, k2, k3, k4 and k5 are rate constants.

Since HBr is formed in steps 'ii' and step 'iii' and disappear in step 'iv', hence, the net rate of formation of HBr-
d[HBr]/dt = k2[H2][Br] + k3[H][Br2] − k4[H][HBr]  ---Eq-1
The H atom are formed in step'ii' and disappear in steps 'iii' and 'iv', hence-
d[H]/dt = k2[Br][H2] − k3[H][Br2] − k4[H][HBr]  ---Eq-2
Applying steady state approximation, we get-
0 = k2[Br][H2] − k3[H][Br2] − k4[H][HBr]
or, k2[Br][H2] = k3[H][Br2] + k4[H][HBr]

The Br atoms are formed in steps 'i', 'iii' and 'iv' and disappear in step 'ii' and 'v', hence-
d[Br]/dt = k1Iabs − k2[Br][H2] + k3[H][Br2] + k4[H][HBr] − k5[Br]2
Applying steady state approximation, we get-
k1Iabs + k3[H][Br2] + k4[H][HBr] = k2[H2][Br] + k5[Br]2  ---Eq-3

Subtracting equation-2 from equation-3 we get-
k1Iabs = k5[Br]2
or, [Br] = (k1Iabs/k5)1/2
Putting the value of [Br] in equation-2 we get-
k2(k1Iabs)1/2[H2] = k3[H][Br2] + k4[H][HBr]
Kinetics of the Formation of HBr

This equation agree with the experimental value. Therefore, the rate of the reaction varies with the square root of the intensity of light(Iabs).

Q. Show that the rate of reaction varies directly to the square root of the intensity of radiation absorbed.


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